If the co-ordinate axes are rotated though though an $\frac{\mathrm{\pi }}{6}$ about the origin then the transformed of equation $\sqrt{3}{x}^2-4xy+\sqrt{3}{y}^2=0$ is

$\begin{array}{r}X=x\cos \theta -y\sin \theta =x\frac{\sqrt{3}}{2}-\frac{y}{2}=\frac{\sqrt{3}x-y}{2}\\ \Rightarrow Y=x\sin \theta +y\cos \theta =\frac{1}{2}x+\frac{\sqrt{3}}{2}y=\frac{x+\sqrt{3}y}{2}\\ Now transformed equation is sub X,Y in given equation \\ we get \sqrt{3}{y}^2-xy=0\\ \end{array}$